I am trying to solve the coin change problem with tail recursion. The recursive solutions I come across are usually something like this
let rec combinations (amount:int) (coins:list<int>) =
if amount = 0 then
1
elif coins.IsEmpty || amount < 0 then
0
else
combinations (amount - coins.Head) coins + combinations amount coins.Tail
clearly inefficient and non tail recursive. I tried to make the solution tail recursive myself:
let combinationsTail (amount:int) (coins:list<int>) : int =
let rec go (amount:int) (sum:int) (coins:list<int>) =
match amount,sum ,coins with
| _,_, [] -> 0
| n,s,_ when n = 0 -> s
| n,_,cs when n < 0 || cs.IsEmpty -> 0
| n,s,h::t -> go (n - h) (n + s) t
go amount 0 coins
But it doesn't work. Does anyone know how to implement a tail recursive solution to this problem? is it even possible?
For achieving tail-recursiveness, you probably want to look into continuation-passing style. Here's an example applied to the Fibonacci sequence, which you could translate verbatim to the coins change problem, since both problems are dealing with aggregation of a recursive tree structure.
It's not the last word on efficiency.
let cc amount coins =
let rec aux k = function
| amount, _ when amount = 0 -> k 1
| amount, _ when amount < 0 -> k 0
| _, [] -> k 0
| amount, hd::tl ->
let k' x =
let k'' y = k (x + y)
aux k'' (amount - hd, hd::tl)
aux k' (amount, tl)
aux id (amount, coins)
Related
needing some help (if possible) in how to count the amount of times a recursive function executes itself.
I don't know how to make some sort of counter in OCaml.
Thanks!
Let's consider a very simple recursive function (not Schroder as I don't want to do homework for you) to calculate Fibonacci numbers.
let rec fib n =
match n with
| 0 | 1 -> 1
| _ when n > 0 -> fib (n - 2) + fib (n - 1)
| _ -> raise (Invalid_argument "Negative values not supported")
Now, if we want to know how many times it's been passed in, we can have it take a call number and return a tuple with that call number updated.
To get each updated call count and pass it along, we explicitly call fib in let bindings. Each time c shadows its previous binding, as we don't need that information.
let rec fib n c =
match n with
| 0 | 1 -> (1, c + 1)
| _ when n > 0 ->
let n', c = fib (n - 1) (c + 1) in
let n'', c = fib (n - 2) (c + 1) in
(n' + n'', c)
| _ -> raise (Invalid_argument "Negative values not supported")
And we can shadow that to not have to explicitly pass 0 on the first call.
let fib n = fib n 0
Now:
utop # fib 5;;
- : int * int = (8, 22)
The same pattern can be applied to the Schroder function you're trying to write.
You can create a reference in any higher scope like so
let counter = ref 0 in
let rec f ... =
counter := !counter + 1;
... (* Function body *)
If the higher scope happens to be the module scope (or file top-level scope) you should omit the in
You can return a tuple (x,y) where y you increment by one for each recursive call. It can be useful if your doing for example a Schroder sequence ;)
I'm teaching myself OCaml, and the main resources I'm using for practice are some problem sets Cornell has made available from their 3110 class. One of the problems is to write a function to reverse an int (i.e: 1234 -> 4321, -1234 -> -4321, 2 -> 2, -10 -> -1 etc).
I have a working solution, but I'm concerned that it isn't exactly idiomatic OCaml:
let rev_int (i : int) : int =
let rec power cnt value =
if value / 10 = 0 then cnt
else power (10 * cnt) (value/10) in
let rec aux pow temp value =
if value <> 0 then aux (pow/10) (temp + (value mod 10 * pow)) (value / 10)
else temp in
aux (power 1 i) 0 i
It works properly in all cases as far as I can tell, but it just seems seriously "un-OCaml" to me, particularly because I'm running through the length of the int twice with two inner-functions. So I'm just wondering whether there's a more "OCaml" way to do this.
I would say, that the following is idiomatic enough.
(* [rev x] returns such value [y] that its decimal representation
is a reverse of decimal representation of [x], e.g.,
[rev 12345 = 54321] *)
let rev n =
let rec loop acc n =
if n = 0 then acc
else loop (acc * 10 + n mod 10) (n / 10) in
loop 0 n
But as Jeffrey said in a comment, your solution is quite idiomatic, although not the nicest one.
Btw, my own style, would be to write like this:
let rev n =
let rec loop acc = function
| 0 -> acc
| n -> loop (acc * 10 + n mod 10) (n / 10) in
loop 0 n
As I prefer pattern matching to if/then/else. But this is a matter of mine personal taste.
I can propose you some way of doing it:
let decompose_int i =
let r = i / 10 in
i - (r * 10) , r
This function allows me to decompose the integer as if I had a list.
For instance 1234 is decomposed into 4 and 123.
Then we reverse it.
let rec rev_int i = match decompose_int i with
| x , 0 -> 10 , x
| h , t ->
let (m,r) = rev_int t in
(10 * m, h * m + r)
The idea here is to return 10, 100, 1000... and so on to know where to place the last digit.
What I wanted to do here is to treat them as I would treat lists, decompose_int being a List.hd and List.tl equivalent.
I am doing some exercises on F#, i have this function that calculate the alternate sum:
let rec altsum = function
| [] -> 0
| [x] -> x
| x0::x1::xs -> x0 - x1 + altsum xs;;
val altsum : int list -> int
The exercise consist in declare the same function with only two clauses...but how to do this?
The answer of mydogisbox is correct and work!
But after some attempts I found a smallest and readable solution of the problem.
let rec altsum2 = function
| [] -> 0
| x0::xs -> x0 - altsum2 xs
Example
altsum2 [1;2;3] essentially do this:
1 - (2 - (3 - 0)
it's is a bit tricky but work!
OFF TOPIC:
Another elegant way to solve the problem, using F# List library is:
let altsum3 list = List.foldBack (fun x acc -> x - acc) list 0;;
After the comment of phoog I started trying to solve the problem with a tail recursive function:
let tail_altsum4 list =
let pl l = List.length l % 2 = 0
let rec rt = function
| ([],acc) -> if pl list then -acc else acc
| (x0::xs,acc) -> rt (xs, x0 - acc)
rt (list,0)
This is also a bit tricky...substraction is not commutative and it's impossible think to revers with List.rev a long list...but I found a workaround! :)
To reduce the number of cases, you need to move your algorithm back closer to the original problem. The problem says to negate alternating values, so that's what your solution should do.
let altsum lst =
let rec altsumRec lst negateNext =
match lst with
| [] -> 0
| head::tail -> (if negateNext then -head else head) + altsumRec tail (not negateNext)
altsumRec lst false
I'm trying to write a function that accepts an int n and returns a list that runs down from n to 0.
This is what I have
let rec downFrom n =
let m = n+1 in
if m = 0 then
[]
else
(m-1) :: downFrom (m - 1);;
The function compiles ok but when I test it with any int it gives me the error
Stack overflow during evaluation (looping recursion?).
I know it's the local varible that gets in the way but I don't know another way to declare it. Thank you!!!
First, the real thing wrong with your program is that you have an infinite loop. Why, because your inductive base case is 0, but you always stay at n! This is because you recurse on m - 1 which is really n + 1 - 1
I'm surprised as to why this compiles, because it doesn't include the rec keyword, which is necessary on recursive functions. To avoid stack overflows in OCaml, you generally switch to a tail recursive style, such as follows:
let downFrom n =
let rec h n acc =
if n = 0 then List.rev acc else h (n-1) (n::acc)
in
h n []
Someone suggested the following edit:
let downFrom n =
let rec h m acc =
if m > n then acc else h (m + 1) (m::acc)
in
h 0 [];
This saves a call to List.rev, I agree.
The key with recursion is that the recursive call has to be a smaller version of the problem. Your recursive call doesn't create a smaller version of the problem. It just repeats the same problem.
You can try with a filtering parameter
syntax:
let f = function
p1 -> expr1
| p2 -> expr2
| p3 -> ...;;
let rec n_to_one =function
0->[]
|n->n::n_to_one (n-1);;
# n_to_one 3;;
- : int list = [3; 2; 1]
As I was writing this function I knew that I wouldn't get tail call optimization. I still haven't come up with a good way of handling this and was hoping someone else might offer suggestions.
I've got a tree:
type Heap<'a> =
| E
| T of int * 'a * Heap<'a> * Heap<'a>
And I want to count how many nodes are in it:
let count h =
let rec count' h acc =
match h with
| E -> 0 + acc
| T(_, value, leftChild, rightChild) ->
let acc = 1 + acc
(count' leftChild acc) + (count' rightChild acc)
count' h 0
This isn't isn't optimized because of the addition of the counts for the child nodes. Any idea of how to make something like this work if the tree has 1 million nodes?
Thanks, Derek
Here is the implementation of count using CPS. It still blew the stack though.
let count h =
let rec count' h acc cont =
match h with
| E -> cont (1 + acc)
| T(_,_,left,right) ->
let f = (fun lc -> count' right lc cont)
count' left acc f
count' h 0 (fun (x: int) -> x)
Maybe I can come up with some way to partition the tree into enough pieces that I can count without blowing the stack?
Someone asked about the code which generates the tree. It is below.
member this.ParallelHeaps threads =
let rand = new Random()
let maxVal = 1000000
let rec heaper i h =
if i < 1 then
h
else
let heap = LeftistHeap.insert (rand.Next(100,2 * maxVal)) h
heaper (i - 1) heap
let heaps = Array.create threads E
printfn "Creating heap of %d elements, with %d threads" maxVal threads
let startTime = DateTime.Now
seq { for i in 0 .. (threads - 1) ->
async { Array.set heaps i (heaper (maxVal / threads) E) }}
|> Async.Parallel
|> Async.RunSynchronously
|> ignore
printfn "Creating %d sub-heaps took %f milliseconds" threads (DateTime.Now - startTime).TotalMilliseconds
let startTime = DateTime.Now
Array.length heaps |> should_ equal threads <| "The size of the heaps array should match the number of threads to process the heaps"
let rec reMerge i h =
match i with
| -1 -> h
| _ ->
printfn "heap[%d].count = %d" i (LeftistHeap.count heaps.[i])
LeftistHeap.merge heaps.[i] (reMerge (i-1) h)
let heap = reMerge (threads-1) E
printfn "Merging %d heaps took %f milliseconds" threads (DateTime.Now - startTime).TotalMilliseconds
printfn "heap min: %d" (LeftistHeap.findMin heap)
LeftistHeap.count heap |> should_ equal maxVal <| "The count of the reMerged heap should equal maxVal"
You can use continuation-passing style (CPS) to solve that problem. See Recursing on Recursion - Continuation Passing by Matthew Podwysocki.
let tree_size_cont tree =
let rec size_acc tree acc cont =
match tree with
| Leaf _ -> cont (1 + acc)
| Node(_, left, right) ->
size_acc left acc (fun left_size ->
size_acc right left_size cont)
size_acc tree 0 (fun x -> x)
Note also that in Debug builds, tail call optimization is disabled. If you don't want to run in Release mode, you can enable the optimization in the project's properties in Visual Studio.
CPS is a good general solution but you might also like to consider explicit use of a stack because it will be faster and is arguably simpler:
let count heap =
let stack = System.Collections.Generic.Stack[heap]
let mutable n = 0
while stack.Count > 0 do
match stack.Pop() with
| E -> ()
| T(_, _, heap1, heap2) ->
n <- n + 1
stack.Push heap1
stack.Push heap2
n